PHP：如何用多個選擇表沖刺？-有解無憂

假設我有來自“testsuite”和“fa”的列“id”。如何使用下面的代碼行查詢“fa”并加入“testsuite”？謝謝。

$query = sprintf("select * from testsuite where start between '%s' and '%s' and suite not like '%s' and suite not like '%s'
                        and serial like '%s'
                        and ipmimac like '%s'
                        and testcase like '%s'
                        order by start desc",
                        mysql_real_escape_string($date1), mysql_real_escape_string($date2),
                        mysql_real_escape_string($lab)
);

以下是我嘗試過的。

$query = sprintf("select * from testsuite where start between '%s' and '%s' and suite not like '%s' and suite not like '%s'
                        and serial like '%s'
                        and ipmimac like '%s'
                        and testcase like '%s'
                        order by start desc",
                        mysql_real_escape_string($date1), mysql_real_escape_string($date2),
                        mysql_real_escape_string($lab),
                        fa.fa as fa.fa_comment
                        FROM fa
                        LEFT JOIN fa.id ON testsuite.fa
);

uj5u.com熱心網友回復：

您在變數之后繼續 SQL 陳述句，因此它不是查詢字串的一部分。此外，JOIN 語法不正確。你可以嘗試類似的東西

"select testsuite.*, fa.fa as fa_comment from testsuite
left join fa ON fa.id = testsuite.fa
where start between '%s' and '%s' and suite not like '%s' and suite not like '%s'
and serial like '%s'
and ipmimac like '%s'
and testcase like '%s'
order by start desc"

此外，您還沒有將足夠的變數傳遞給 sprintf。他們需要匹配%s- 但是我強烈建議不要使用 sprint 和 mysql_real_escape_string，而是依賴使用 PHP 中可用的 mysqli 或 PDO 包的準備好的陳述句。

uj5u.com熱心網友回復：

您必須將連接放在 SQL 字串中，而不是sprintf.

$query = sprintf("
    select *
    from testsuite AS ts
    LEFT JOIN fa ON fa.id = ts.id
    where start between '%s' and '%s' 
            and suite not like '%s'
            and serial like '%s'
            and ipmimac like '%s'
            and testcase like '%s'
    order by start desc",
                 mysql_real_escape_string($date1), mysql_real_escape_string($date2),
                 mysql_real_escape_string($lab)
);

uj5u.com熱心網友回復：

我可以在這里看到很多問題。我不知道我從哪里開始...

但是，我認為您搜索類似的內容：

$query = sprintf("
SELECT
    testsuite.*,
    fa.fa,
    fa.fa_comment
FROM testsuite
LEFT JOIN fa
    ON fa.id = testsuite.fa
WHERE
    testsuite.start BETWEEN '%s' AND '%s'
    AND testsuite.suite NOT LIKE '%s'
    AND testsuite.suite NOT LIKE '%s'
    AND testsuite.serial LIKE '%s'
    AND testsuite.ipmimac LIKE '%s'
    AND testsuite.testcase LIKE '%s'
ORDER BY
    testsuite.start DESC",
                        mysql_real_escape_string($date1), mysql_real_escape_string($date2),
                        mysql_real_escape_string($lab)
);

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標籤：phpsql

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